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Is the integer N odd? (1) N is divisible by 3 (2)2N is [#permalink]
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23 Oct 2006, 10:58
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Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.



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Joined: 05 Jul 2006
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Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.
from one n could be 6 or 9 ( even or odd)...not suff
from two
assume n = 3 devisible by 3,1 , 2n = 6 devisble by 6,3,2,1
assume n = 5 devisble by 5,2 2n = 10 devisble by 10,5,2,1
assume n = 9 devisble by 3 , 9 ,1 while 18 is devisble by 2,9,3,2,6,1
suff and n is odd
My answer is B



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Re: DS  good concept [#permalink]
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23 Oct 2006, 22:13
I think B too.
1 is not sufficient. N can be either odd or even.
For 2: All the prime factors of an odd number N will be odd. The prime factors to 2N will be all the factors of N as well as the factors of N multiplied by 2.
For example: 21 = 1,3,7,21
42 = 1,3,7,21,
2,6,14,42 (above factors multiplied by 2)
For an even number N, 2N can't have all distinct prime factors by multiplying N's factors by 2
10  1,2,5,10
20  1,2,5,10
4, ,20



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Joined: 10 Oct 2005
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yezz wrote: Is the integer N odd? (1) N is divisible by 3 (2)2N is divisible by twice as many positive integer as N.
from one n could be 6 or 9 ( even or odd)...not suff
from two
assume n = 3 devisible by 3,1 , 2n = 6 devisble by 6,3,2,1
assume n = 5 devisble by 5,2 2n = 10 devisble by 10,5,2,1
assume n = 9 devisble by 3 , 9 ,1 while 18 is devisble by 2,9,3,2,6,1
suff and n is odd
My answer is B
IMHO E it is
yezz and what if n=4?
4 is devisible by 4,2,1 and 8 is divisible by 8, 4, 2,1 n is even
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Joined: 01 Oct 2006
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Yurik79, I did not understand how your example satisfies the 2nd option.
8 is divisible by 4 positive numbers and 4 is divisible by 3 numbers. 8 should have 6 positive factors to meet the requirement stated in 2. Am I missing something?



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Joined: 01 Oct 2006
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I tried solving it in another way:
Let O be any odd number. O is prime factorized into a, b and c.
O = a^m x b^n x c^l
Therefore number of factors of O = (m+1)(n+1)(l+1)
2xO = 2^1 x a^m x b^n x c^l
Therefore number of factors of 2O = (1+1)(m+1)(n+1)(l+1); this is twice the number of factors of O.
Now let E be any even number.
E= 2^m x d^n x y^l
Therefore number of factors of E = (m+1)(n+1)(l+1)
// At the minimum m+1 = 2
2xE = 2^(m+1) x d^n x y^l
Therefore number of factors of 2E = (m+2)(n+1)(l+1)
The number of prime factors of 2E will be twice of those of E only if (m+2) = 2(m+1)
This is possible only if m=0. Since E is an even number, m cannot be 0.



Current Student
Joined: 29 Jan 2005
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agreed on (B). Statement 1 will always result in just one more postive factor.



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mst wrote: Yurik79, I did not understand how your example satisfies the 2nd option. 8 is divisible by 4 positive numbers and 4 is divisible by 3 numbers. 8 should have 6 positive factors to meet the requirement stated in 2. Am I missing something?
my fault))you are absolutely correct thanks for pointing my mistake
yezz I am sorry agree B
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